phan tich cac da thuc sau thanh nhan tu
a, (x-y)3 - (z-y)3 - (x-z)3
b, (2016x-2015)3 + (2014-2013x)3 + ( 1-3x)3
c, (2a+b+c)3 - (a-b+2c)3 - (a+2b-c)3
phan tich da thuc sau thanh nhan tu:
a) (3x-2)(4x-3)-(2-3x)(x-1)-2(3x-2)(x+1)
b) x^2(y-z)+y^2(z-x)+z^2(x-y)
phan tich da thuc thanh nhan tu (x-y).z^3 +(y-z).x^3 +(z-y).y^3
(x+y+z)^3 - (x+y-z)^3 - (y+z-x)^3 - (z+x-y)^3
Phan tich da thuc thanh nhan tu
Goi da thuc tren la A
Thay a=b -> A= 0 -> A chua nghiem la a-b
Tuong tu b=c-> A = 0 - > A chua nghiem la b -c
Tuong tu c =a - > A = 0 -> A chua nghiem la c-a
=> A = k(a - b)(b - c)(c - a)
Vì A có bậc 3 mà (a - b)(b - c)(c - a) cũng có bậc 3 -> k là 1 số
Thay a = 3, b= 2, c= 1
=> A= -6=k.1.1..-2
=> k = 3
=> A = 3(a - b)(b - c)(c - a)
Đây gọi là phương pháp giá trị riêng bạn nha!
x^5 + x + 1
= x^5 - x^2 + (x^2 + x + 1)
= x^2(x^3 - 1) + ( x^2 + x + 1)
= x^2( x - 1)(x^2 + x + 1) + ( x^2 + x + 1)
= (x^3 - x^2 + 1)(x^ 2 + x + 1)
PHAN TICH DA THUC THANH NHAN TU
B= X^3+Y^3+Z^3-3XYZ
\(B=x^3+y^3+z^3-3xyz=\left(x+y\right)^3+z^3-3xy\left(x+y\right)-3xyz\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2+2xy-yz-xz\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
phan tich da thuc thanh nhan tu
( a + b + c)3 - a3 - b3 - c3
8( x + y + z)3 - ( x+ y )3 - ( y + z )3 - ( z + x)3
a) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
tk mình nhoa bạn, mình kp vs bn r đó
Phan tich da thuc thanh nhan tu :
a, -x^3 * ( 2x + 1 )^2 + 49x
b, 125x^2 + 20y - 5y^2 - 20
c, ( 1+ 2x )*(1 - 2x ) - x*(x + 2 )*( x - 2 )
d, ( x - z )*(x + z) - y*(2x - y )
e, x^2 - 3x - 54
Giup minh voi nhe cac ban. Minh se tick cho.
a: \(=x\left[49-x^2\left(2x+1\right)^2\right]\)
\(=x\left[49-\left(2x^2+x\right)^2\right]\)
\(=x\left[\left(7-2x^2-x\right)\left(7+2x^2+x\right)\right]\)
b: \(=5\left[25x^2-\left(y^2-4y+4\right)\right]\)
\(=5\left[\left(5x-y+2\right)\left(5x+y-2\right)\right]\)
c: \(=1-4x^2-x\left(x^2-4\right)\)
\(=1-4x^2-x^3+4x\)
\(=\left(1-x\right)\left(1+x+x^2\right)-4x\left(x-1\right)\)
\(=\left(1-x\right)\left(1+x+x^2+4x\right)\)
\(=\left(1-x\right)\left(x^2+5x+1\right)\)
e: =(x-9)(x+6)
BAI 1.phan tich cac da thuc sau thanh nhan tu:
a,2x^2-2xy-5x+5y
b,8x^2+4xy-2ax-ay
c,x^3-4x^2+4x
d,2xy-x^2-y^2+16
e,x^2-y^2-2yz-z^2
g,3a^2-6ab+3b^2-12c^2
BAI 2.tinh nhanh
a,37,5.8,5-7,5.3,4-6,6.7,5+1,5.37,5
b,35^2+40^2-25^2+80.35
BAI 3. Tim x biet:
a,x^3-1/9x=0
b,2x-2y-x^2+2xy-y^2=0
c,x(x-3)+x-3=0
d,x^2(x-3)+27-9x=0
BAI 4.Phan tich cac da thuc sau thanh nhan tu
a,x^2-4x+3
goi y :tach-4x=-x3xhoac tach3=-1+4
b,x^2+x-6
c,x^2-5x+6
d,x^4+4 (goi y:them va bot 4x^2)
BAI 5.Chung minh rang;
(3n+4)^2-16 chia het cho 3 voi moi so nguyen n.
BAI 6.Tinh gia tri cua bieu thuc sau:
M=a^3-a^2b-ab^2+b^3 voi a=5,75:b=4,25
BAI 7.Tim x biet:
a,x^2+x=6
b,6x^3+x^2=2x
Bài 1 câu g bạn kia làm sai mình sửa lại nhá
\(3a^2-6ab+3b^2-12c^2\)
\(=3\left(a^2-2ab+b^2\right)-12c^2\)
\(=3\left(a-b\right)^2-12c^2\)
\(=3\left[\left(a-b\right)^2-4c^2\right]\)
\(=3\left(a-b-2c\right)\left(a-b+2c\right)\)
Để mình làm tiếp cho :))
Bài 2 :
Câu a : \(37,5.8,5-7,5.3,4-6,6.7,5+1,5.37,5\)
\(=\left(37,5.8,5+1,5.37,5\right)-\left(7,5.3,4+6,6.7,5\right)\)
\(=37,5\left(8,5+1,5\right)-7,5\left(3,4+6,6\right)\)
\(=37,5.10-7,5.10\)
\(=10.30=300\)
Câu b : \(35^2+40^2-25^2+80.35\)
\(=\left(35^2+80.35+40^2\right)-25^2\)
\(=\left(30+45\right)^2-25^2\)
\(=75^2-25^2\)
\(=\left(75+25\right)\left(75-25\right)\)
\(=100.50=5000\)
Bài 3 :
Câu a : \(x^3-\dfrac{1}{9}x=0\)
\(\Leftrightarrow x\left(x^2-\dfrac{1}{9}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-\dfrac{1}{9}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\pm\dfrac{1}{3}\end{matrix}\right.\)
Câu b : \(2x-2y-x^2+2xy-y^2=0\)
\(\Leftrightarrow2\left(x-y\right)-\left(x-y\right)^2=0\)
\(\Leftrightarrow\left(x-y\right)\left(2-x+y\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-y=0\\2-x+y=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=y\\x+y=2\Rightarrow x=2-y\end{matrix}\right.\)
Câu c :
\(x\left(x-3\right)+x-3=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
\(x^2\left(x-3\right)+27-9x=0\)
\(\Leftrightarrow x^2\left(x-3\right)-9\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x^2-9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=\pm3\end{matrix}\right.\)
Bài 4 :
Câu a :
\(x^2-4x+3\)
\(=x^2-x-3x+3\)
\(=\left(x^2-x\right)-\left(3x-3\right)\)
\(=x\left(x-1\right)-3\left(x-1\right)\)
\(=\left(x-1\right)\left(x-3\right)\)
Câu b :
\(x^2+x-6\)
\(=x^2-2x+3x-6\)
\(=x\left(x-2\right)+3\left(x-2\right)\)
\(=\left(x-2\right)\left(x+3\right)\)
Câu c :
\(x^2-5x+6\)
\(=x^2-2x-3x+6\)
\(=\left(x^2-2x\right)-\left(3x-6\right)\)
\(=x\left(x-2\right)-3\left(x-2\right)\)
\(=\left(x-2\right)\left(x-3\right)\)
Câu d :
\(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)
Bài 1:
a) \(2x^2-2xy-5x+5y\)
\(=\left(2x^2-2xy\right)-\left(5x-5y\right)\)
\(=2x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(2x-5\right)\)
b) \(8x^2+4xy-2ax-ay\)
\(=\left(8x^2+4xy\right)-\left(2ax+ay\right)\)
\(=4x\left(2x+y\right)-a\left(2x+y\right)\)
\(=\left(2x+y\right)\left(4x-a\right)\)
c) \(x^3-4x^2+4x\)
\(=x\left(x^2-4x+4\right)\)
\(=x\left(x-2\right)^2\)
d) \(2xy-x^2-y^2+16\)
\(=-\left[\left(x^2-2xy+y^2\right)-16\right]\)
\(=-\left[\left(x-y\right)^2-4^2\right]\)
\(=-\left[\left(x-y-4\right)\left(x-y+4\right)\right]\)
e) \(x^2-y^2-2yz-z^2\)
\(=-\left[\left(z^2+2yz+y^2\right)-x^2\right]\)
\(=-\left[\left(z+y\right)^2-x^2\right]\)
\(=-\left[\left(z+y+x\right)\left(z+y-x\right)\right]\)
g) \(3a^2-6ab+3b^2-12c^2\)
\(=\left(3a^2-6ab+3b^2\right)-12c^2\)
\(=\left(\sqrt{3a}+\sqrt{3b}\right)^2-12c^2\)
\(=\left(\sqrt{3a}+\sqrt{3b}+\sqrt{12c}\right)\left(\sqrt{3a}+\sqrt{3b}-\sqrt{12c}\right)\)
phan tich da thuc thanh nhan tu x^3 +y^3-z^3+3xyz
\(x^3+y^3+z^3-3xyz\) \(=\left(x+y\right)^3-3x^2y-3xy^2+z^2-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
HỌC TỐT NHA!
ta có:
x³ + y³ + z³ - 3xyz
= (x+y)³ - 3xy(x-y) + z³ - 3xyz
= [(x+y)³ + z³] - 3xy(x+y+z)
= (x+y+z)³ - 3z(x+y)(x+y+z) - 3xy(x-y-z)
= (x+y+z)[(x+y+z)² - 3z(x+y) - 3xy]
= (x+y+z)(x² + y² + z² + 2xy + 2xz + 2yz - 3xz - 3yz - 3xy)
= (x+y+z)(x² + y² + z² - xy - xz - yz)
- Hình như đề của u sai hay sao á :)))
Phan tich da thuc thanh nhan tu
a(a+2b)^3-b(2a+b)^3
=a(a+2b)^3-[-b(a+2b)^3]
=(a+2b)^3(a+b)